Motivation.

Consider a cylindrical distribution of mass (or charge) as in figure (\ref{fig:cylinderPotential:cylinderPotentialFig1}), with points in the cylinder given by coordinates, and the point of measurement of the potential measured at .

Attempting to evaluate the integrals.

With

This is found to be

It is clear that we can’t evaluate this limit directly for since that gives us in the logarithm term. Presuming this can be evaluated, we must have to evaluate the complete set of integrals first, then take the limit. Based on the paper http://www.ifi.unicamp.br/~assis/J-Electrostatics-V63-p1115-1131(2005).pdf, it appears that this can be evaluated, however, the approach used therein uses mathematics a great deal more sophisticated than I can grasp without a lot of study.

Can we proceed blindly using computational tools to do the work? Attempting to evaluate the remaining integrals with Mathematica fails, since evaluation of both

and

either time out, or take long enough that I aborted the attempt to let them evaluate.

Alternate evaluation order?

We can also attempt to evaluate this by integrating in different orders. We can for example do the coordinate integral first

It should be noted that this returns a number of hard to comprehend ConditionalExpression terms, so care manipulating this expression may also be required.

If we try the angular integral first, we get

where

is the complete elliptic integral of the first kind. Actually evaluating this integral, especially in the limiting case, probably requires stepping back and thinking a bit (or a lot) instead of blindly trying to evaluate.

Motivation.

Part of a classical mechanics problem set was to look at what portions of momentum and angular momentum are conserved for various fields. Since this was also a previous midterm question, I’m expecting that some intuition was expected to be used determine the form of the Lagrangians, with not much effort on finding the precise form of the potentials. Here I try to calculate one such potential explicitly.

Oddly, I can explicitly calculate the potential for the infinite homogeneous plane if I start with the force and then calculate the potential, but if I start with the potential the integral also diverges? That doesn’t make any sense, so I’m wondering if I’ve miscalculated. I’ve tried a few different ways below, but can’t get a non-divergent result if I start from the integral definition of the potential instead of deriving the potential from the force after adding up all the directional force contributions (where I find of course that all the component but the one perpendicular to the plane cancel out).

Forces and potential for an infinite homogeneous plane.

Calculating the potential from the force.

For the plane, with as the distance from the plane, and coordinates in the plane, as illustrated in figure (\ref{fig:infiniteSheetPotentials:infiniteSheetPotentialsFig1}). The gravitational force from an element of mass on the plane is
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{infiniteSheetPotentialsFig1}
\caption{Coordinate choice for interaction with infinite plane mass distribution.}
\end{figure}

Integrating for the total force on the test mass, noting that the sinusoidal terms vanish when integrated over a interval, we have

a substitution gives us

For

For the Lagrangian problem we want the potential

or

This is a reasonable seeming answer. Our potential is of the form

with

Calculating the potential directly.

If we only want the potential, why start with the force? We ought to be able to work with potentials directly, and write

It’s a quick calculation to verify that is correct, and we find (provided )

To verify the signs, it’s helpful to refer to the diagram (\ref{fig:infiniteSheetPotentials:infiniteSheetPotentialsFig2}) which illustrates the position vectors. Now, suppose we calculate the potential directly
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{infiniteSheetPotentialsFig2}
\caption{Direction vectors for interaction with mass distribution.}
\end{figure}

Naive approach, with bogus result.

Our mass density as a function of is

Again we can make a tangent substitution

for

This has the same functional form as 2.4, except with

There’s one significant and irritating difference. The integral above doesn’t have a unit value, but instead diverges (with as ).

What went wrong? Trouble can be seen right from the beginning. Consider the differential form above for

This we are integrating on , so long before we make the trig substitutions we are in trouble.

As the limit of a finite volume.

My guess is that we have to tie the limits of the width of the plane and its diameter, decreasing the thickness in proportion to the increase in the radius. Let’s try that.

With a finite cylinder of height , radius , with a measurement of the potential directly above the cylinder at height , we have

Performing the integration and substituting

we have

For this integral we find

Taking limits and the terms at either or tend to 1. This leaves us only with the contribution, so

With we have the structure of a differential above, but instead of expressing the derivative in the usual forward difference form

we have to use a backwards difference, which is equivalent, provided the function is continuous at

We can then form the differential

We also want to express the charge density in terms of surface charge density , and note that these are related by .

or

Looking at this result, we have the same divergent integration result as in the first attempt, and the reason for this is clear after some reflection. The limiting process for the radius and the thickness of the slice were allowed to complete independently. Before taking limits we had

Consider a similar, but slightly more general case, where we evaluate the limit

where , so that

So even without evaluating the integral we expect that we’ll have

This matches what was obtained in 2.19 by brute forcing the integral with Mathematica, and then evaluating the limit the hard way. Darn.

As the limit of a finite volume. Take II.

Now we’ll reduce the thickness of the volume, keeping the total mass fixed, so that

or

We wish to evaluate

Performing the integrals, (first , then ) we find

The term is clearly killed in the limit, and we have a contribution from the first term. For the difference of logarithms above can be written as

Suppose we could also validly argue that this tends to , and the difference of square roots could also be canceled. Then we would be left with just

If we also demand that , then we have a contribution from the remaining square root and are left with

This differs from the expected result by a factor of , and we’ve had to do some very fishy root taking to even get that far. Employing l’H\^opital’s rule (or letting Mathematica attempt to evaluate the limit), we get infinities for the difference of logarithm term.

So, with a lot of cheating we get a result that’s similar to the expected, but not actually a match, and even to get that we had to take the limits in an invalid way. It looks like it’s back to the drawing board, but I’m not sure how to approach it.

After thinking about it a bit, perhaps the limiting process for the width needs to be explicitly accounted for using a delta function? Perhaps a QM like treatment where we express the integral in terms of some basis and look for the resolution of identity in that resolution?

On UNIX there’s a one to one mapping between instance owner userid and the path to the diaglog, and unless you’ve redirected it, you’ll find it in ~/sqllib/db2dump/. Here’s a reminder for myself of the magic required to find your db2diag.log file in a Windows instance:

I think this test case must have been written assuming that the ulimit for number of file handles was some nice small number. This does a nice number on the stack, and was fairly non-obvious when looking in the debugger session.

The debugger shows me the test function (testgroup2) driving a trap in a called function:

where my trap is on a dereference to a variable gtfCBInfo that was good earlier in the code.

if ( gtfCBInfo->logptr == NULL )

Yet, it was clear that there were no intervening assignments to gtfCBInfo. I was curious exactly how this corruption propagated once found, and that’s clear enough once you look at the assembly for a function call immediately after the corruption. See all the code that sets up the parameters for the call by fetching stack spills (the loads from (%rbp)):

Because we’ve walked all over the neighbourhood of 0(%rbp ) with the open() calls, once we try to fetch them we start passing garbage. It’s so common to see stack corruptions manifest in even nastier ways (like a trap on return from a function call because we’ve wiped out the return address), so I was a bit suprised to see such a run of the mill seeming corruption here as a result. Perhaps if we hadn’t called any functions, we would have trapped on unwind instead?

Sometimes we can utilize solutions already found to understand the behaviour of more complex systems. Combining the two we can look at flow over a plate as in figure (\ref{fig:continuumL12:continuumL12fig3})

Example 2. Fluid in a container. If the surface tension is altered on one side, we induce a flow on the surface, leading to a circulation flow. This can be done for example, by introducing a heat source or addition of surfactant.

This sort of flow is hard to analyze, only first done by Steve Davis in the 1980’s. The point here is that we can use some level of intuition to guide our attempts at solution.

Flow down a pipe.

Reading: section 2 from [1].

Recall that the Navier-Stokes equation is

We need to express this in cylindrical coordinates as in figure (\ref{fig:continuumL12:continuumL12fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig5}
\caption{Flow through a pipe.}
\end{figure}

Our gradient is

For our Laplacian we find

which we can write as

NS takes the form

For steady state and incompressible fluids in the absence of body forces we have

or, in coordinates

With an assumption that we have no radial or circulatory flows (), and with assumed to only have a radial dependence, our velocity is

and an assumption of linear pressure dependence

then NS takes the final simple form

Solving this we have

Requiring finite solutions for means that we must have . Also , we have so we must have

and seek simultaneous solutions to the pair of stress tensor equations

In general this requires an iterated approach, solving for one with an initial approximation of the other, then switching and tuning the numerical method carefully for convergence.

We expect that the flow of liquid will induce a flow of air at the interface, but may be able to make a one-sided approximation. Let’s see how far we get before we have to introduce any approximations and compute the traction vector for the liquid

So

Our boundary value condition is therefore

When can we decouple this, treating only the liquid? Observe that we have

so if

we can treat only the liquid portion of the problem, with a boundary value condition

Let’s look at the component of the traction vector in the direction of the normal (liquid pressure acting on the air)

or

i.e. We have pressure matching at the interface.

Our body force is

Referring to the Navier-Stokes equation 4.4, we see that our only surviving parts are

It’s important to note that in these problems we have to derive our boundary value conditions! They are not given.

In this discussion, the height was assumed to be constant, with the tangential direction constant and parallel to the surface that the liquid is flowing on. It’s claimed in class that this is actually a consequence of surface tension only! That’s not at all intuitive, but will be covered when we learn about “stability conditions”.

Study note.

Memorizing the NS equation is required for midterm, but more complex stuff (like cylindrical forms of the strain tensor if required) will be given.

Disclaimer.

Navier-Stokes equation.

The Navier-Stokes equation (our fluids equivalent to Newton’s second law) was found to be

In this course we’ll focus on the incompressible case where we have

We watched a video of the rocking tank as in figure (\ref{fig:continuumL11:continuumL11fig1}). The boundary condition that accounted for the matching of the die marker is that we have no slipping at the interface. Writing for the tangent to the interface then this condition at the interface is described mathematically by the conditions

Steady incompressible rectilinear (unidirectional) flow.

We start with the incompressibility condition, which written explicitly, is

or

This implies

so our velocity can only be function of the and coordinates only

The non-linear term of the Navier-Stokes equation takes the form

With incompressibility and conditions killing this term, and the steady state condition 3.13 killing the term, the Navier-Stokes equation for this incompressible unidirectional steady state flow (in the absence of body forces) is reduced to

Example: Shearing flow.

The flows of this sort don’t have to be trivial. For example, even with constant pressure () as in figure (\ref{fig:continuumL11:continuumL11fig4}) we can have a “shearing flow” where the fluids at the top surface are not necessarily moving at the same rates as the fluid below that surface. We have fluid flow in the direction only, and our velocity is a function only of the coordinate.

Example: Channel flow

This time our simplified Navier-Stokes equation 3.19a is reduced to something slightly more complicated

with solution

The boundary value conditions with the coordinate system in use illustrated in figure (\ref{fig:continuumL11:continuumL11fig5}) require the velocity to be zero at the interface (the pipe walls preventing flow in the interior of the pipe)

It’s clear that this is maximized by , but we can also see this by computing

This maximum is

The flux, or flow rate is

Let’s now compute the strain () and the stress ()

stress

This can be used to compute the forces on the inner surfaces of the tube. As illustrated in figure (\ref{fig:continuumL11:continuumL11fig7}), our normals at are respectively. The traction vector in the direction is at is

FIXME: ask in class. This is the tangential force at the boundary of the wall. What is it a force on? If it is tangential, how can it act on the wall? It could act on an impediment placed right up next to the wall, but if that’s the case, why are we integrating from to ?

In Lotus Notes version 8, once an email is sent with an incorrect email address, it can have an almost viral effect. For example, we have a couple internal distribution lists that have been used incorrectly a few times:

I’m on one such distribution list, and this incorrect address has been repeatedly added to my local address book, so once I try to send an email to our askOSS list, it bounces.

The workaround is to search through the local address book, find the bad addresses, and delete them. This unfortunately has to be done on any replicas separately, and it’s temporary because somebody ends up not knowing to do this, and the bad address ends up back in everybodies local address books again.

I asked our IT support folks about this. They’ve opened a defect against Lotus notes for this issue, but have also provided the following workaround:

As a temporary fix

You will have to delete the incorrect address from your address book under recent contacts which you have been doing

create a contact in the address book with the incorrect address as the main contact name with the right mail address in the contents of it

you can disable the recent contact feature to add the names to the recent contacts field (before doing this you will have clear all the bad address from Contacts|recent contacts field)

Steps to disable recent contacts:

Click File|preferences|contacts

Check “Do not automatically add contacts to the recent contacts view”

click ok

I’m hopeful that this will deal with the problem. The side effect will be that I will have a smaller contact list, but I think I can live with that.

Motivation.

Notes from Prof. Poppitz’s phy354 classical mechanics lecture on the Runge-Lenz vector, a less well known conserved quantity for the 3D potentials that can be used to solve the Kepler problem.

Motivation: The Kepler problem.

We can plug away at the Lagrangian in cylindrical coordinates and find eventually

but this can be messy to solve, where we get elliptic integrals or worse, depending on the potential.

For the special case of the 3D problem where the potential has a form, this is what Prof. Poppitz called “super-integrable”. With conserved quantities to be found, we’ve got one more. Here the form of that last conserved quantity is given, called the Runge-Lenz vector, and we verify that it is conserved.

Runge-Lenz vector

Given a potential

and a Lagrangian

and writing the angular momentum as

the Runge-Lenz vector

is a conserved quantity.

Verify the conservation assumption.

Let’s show that the conservation assumption is correct

Here, we note that angular momentum conservation is really , so we are left with only the acceleration term, which we can rewrite in terms of the Euler-Lagrange equation

We can compute the double cross product

For

Plugging this we have

Now let’s look at the other term. We’ll need the derivative of

Putting all the bits together we’ve now verified the conservation statement

With

our vector must be some constant vector. Let’s write this

so that

Dotting 3.12 with we find

With lying in the plane of the trajectory (perpendicular to ), we must also have lying in the plane of the trajectory.

Now we can dot 3.12 with to find

This is

This is a kind of curious implicit relationship, since is also a function of . Recall that the kinetic portion of our Lagrangian was

so that our angular momentum was

with no dependence in the Lagrangian we have

or

Our dynamics are now fully specified, even if this not completely explicit

What we can do is rearrange and separate variables

to find

Now, at least is specified implicitly.

We can also use the first of these to determine the magnitude of the radial velocity

with this, we can also find the energy

Or

Is this what was used in class to state the relation

It’s not obvious exactly how that is obtained, but we can go back to 3.18 to eliminate the term

Presumably this simplifies to the desired result (or there’s other errors made in that prevent that).